Problem: If $x \otimes y = 2x-4$ and $x \diamond y = x^{2}+4y^{2}$, find $-2 \otimes (-4 \diamond -3)$.
We don't need to find $-4 \diamond -3$ because $x \otimes y$ depends only on the left operand. Find $-2 \otimes y$ $ -2 \otimes y = (2)(-2)-4$ $ \hphantom{-2 \otimes y} = -8$.